For $a,b \in \mathbb{R}$, prove that $ ab \le a^2 + b^2$

Question

Just what the title says. Im a bit stuck here.

Edit:I'm thinking that since a and b are both reals, they can be either positive or negative, making ab either positive or negative. I also know that $a^2≥0$ as well as $b^2≥0$. I'm just have trouble making formalizing it.

In my head, I'm thinking that since a and b are both reals, they can be either positive or negative, making ab either positive or negative. I also know that $a^2 \ge 0$ as well as $b^2 \ge 0$. I'm just have trouble making formalizing it. — kfox25, Jun 25 '20 at 01:05

score 1 · Accepted Answer · answered Jun 25 '20 at 01:09

1

First approach

\begin{align*} a^{2} + b^{2} \geq ab & \Longleftrightarrow 2(a^{2} + b^{2}) \geq 2ab\\\\ & \Longleftrightarrow (a^{2} - 2ab + b^{2}) + (a^{2} + b^{2}) \geq 0\\\\ & \Longleftrightarrow (a-b)^{2} + (a^{2} + b^{2}) \geq 0 \end{align*}

Second approach

\begin{align*} a^{2} + b^{2} \geq ab & \Longleftrightarrow a^{2} - ab + b^{2} \geq 0\\\\ & \Longleftrightarrow \left(a - \frac{b}{2}\right)^{2} + \frac{3b^{2}}{4} \geq 0 \end{align*}

answered Jun 25 '20 at 01:09

user0102

21,572

On the second line, how was $a^2 + b^2$ kept? I see how $2ab$ was moved over but what happened to the $2(a^2 + b^2)?$. (Sorry for the bother) – kfox25 Jun 25 '20 at 01:15
1

Notice that $2(a^{2} + b^{2}) = (a^{2} + b^{2}) + (a^{2} + b^{2})$. Then I've just gathered the first term with $-2ab$. Can you take it from there? – user0102 Jun 25 '20 at 01:16
I see now. Thanks so much for your help! – kfox25 Jun 25 '20 at 01:17
You're welcome. – user0102 Jun 25 '20 at 01:17

For $a,b \in \mathbb{R}$, prove that $ ab \le a^2 + b^2$

1 Answers1